Problem: Let $f(x)=\log(x^2-1)$. Find $f'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{2x}{x^2-1}$ (Choice B) B $\dfrac{2x}{(x^2-1)\ln(10)}$ (Choice C) C $\dfrac{1}{(x^2-1)\ln(10)}$ (Choice D) D $\dfrac{x^2-1}{(x)\ln(10)}$
$f$ is a logarithmic function, but its argument isn't simply $x$. Therefore, it's a composite function. In other words, suppose $u(x)=x^2-1$, then $f(x)=\log\Bigl(u(x)\Bigr)$. $f'(x)$ can be found using the following identity: $\dfrac{d}{dx}\left[\log\Bigl(u(x)\Bigr)\right]=\dfrac{u'(x)}{\ln(10)u(x)}$ [Why is this identity true?] Let's differentiate! $\begin{aligned} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}\log(x^2-1) \\\\ &=\dfrac{d}{dx}\log\Bigl(u(x)\Bigr)&&\gray{\text{Let }u(x)=x^2-1} \\\\ &=\dfrac{u'(x)}{\ln(10)u(x)} \\\\ &=\dfrac{2x}{\ln(10)(x^2-1)}&&\gray{\text{Substitute }u(x)\text{ back}} \\\\ &=\dfrac{2x}{(x^2-1)\ln(10)} \end{aligned}$ In conclusion, $f'(x)=\dfrac{2x}{(x^2-1)\ln(10)}$.